**Course Links: goal 2017 junior Math League training Team-1 (Zhao Yin)**

**1, right Angle $\triangle{abc}$, $AD $ is the hypotenuse on the high, $I _1, i_2$ is $\triangle{abd}, \triangle{acd}$ heart, verify: $B, C, I_2, i_1$ four points round.**

**Prove:**

Consider proof $$\angle{bi_1i_2} + \angle{bci_2} = \angle{bi_1d} + \angle{di_1i_2} + \angle{bci_2} = 180^\circ.$$ easy to know $$\angle{BI_1D} = 180^\circ-{1\over2} (\angle{abd} + \angle{adb}) = 180^\circ-{1\over2} (180^\circ-\angle{bad}) =90^\circ + {1\over2} \angle{bad}.$$ on the other hand, $$\angle{bac} = \angle{i_1di_2} = 90^\circ,\ Ab:ac = i_1d:i_2d\ (\because\triangle{abd} \sim \tri ANGLE{CAD}). $$ so $$\triangle{abc} \sim \triangle{di_1i_2}\rightarrow \angle{abc} = \angle{di_1i_2},\ \angle{ACB} = \ angle{di_2i_1}.$$ this can be $$\angle{bi_1i_2} + \angle{bci_2} = \angle{bi_1d} + \angle{di_1i_2} + \angle{bci_2}$$ $$= 90^\circ + {1\over2}\angle{bad}+ \angle{abc} + {1\over2}\angle{acb}$$ $$= 90^\circ + {1\over2}\angle{bad}+ \angle{ABC} + {1\over2} \angle{bad}$$ $$= 90^\circ + \angle{bad}+ \angle{abc} = 180^\circ.$$

**Q.E.D.**

**2, in the convex Pentagon $ABCDE $, if $\angle{abc} = \angle{ade}$, $\ANGLE{AEC} = \angle{adb}$, verify: $\angle{bac} = \angle{dae}$.**

**Prove:**

Set $BD, the ce$ intersection is $F $. by $\ANGLE{AEC} = \angle{adb}$ $A, E, D, f$ Four points are all round.

Again by $\angle{abc} = \angle{ade} = \angle{afe}$ know $A, B, C, f$ Four points are all round.

This can be $\angle{bac} = \ANGLE{BFC} = \angle{dfe} = \angle{dae}$.

**Q.E.D.**

**3, in the circle of the string $AB, ac$, $\angle{bac}$ of the split line round at the point $D $. Over $D $ for $DE \perp ab$ in $E $, verify: $AE = \displaystyle{1\over 2} (AB + AC) $.**

**Prove:**

Consider proving $2ae = AB + ac$.

By the nature of the angular line and the $CD = bd$, constructs the congruent triangle, makes $DF \perp ac$ (or its extension cord) in $F $.

Easy to $\triangle{dfc}\cong\triangle{deb}$ and $\triangle{dfa}\cong\triangle{dea}$, so $2ae = AE + AF = AB + ac$.

It should be noted that the case should be explained when the $B, e$ coincident. Ishi at this time $AD $ for the diameter of the circle, the conclusion is still established.

**Q.E.D.**

**4, in the Triangle $ABC $, $\ANGLE{ACB} = 45^\circ$, $D $ for $AC $ on a point and $\angle{adb} = 60^\circ$, $AB $ cut $\triangle{bcd}$ circumscribed circle in $B $, verify: $AD:D C = 2:1$.**

**Prove:**

Because of the $AD, the dc$ in the same line, considering the parallel cut theorem and the parallel relationship to solve.

Connect $OB $ to create a right angle, and then by $\ANGLE{ACB} = 45^\circ$ Consider recreating a $90^\circ$: $$\angle{bod} = 2\ANGLE{BCD} = 90^\circ.$$ thus found $AB \parallel Od\rightarrow AD:DC = be:ec$. Next you can consider proving $BE: EC = 2:1$.

Calculated by $\triangle{abc}\sim\triangle{adb}$ and simple angle: $$\angle{abc} = \angle{adb} = 60^\circ\rightarrow \angle{OBC} = 30^\ Circ\rightarrow be = 2oe,$$ $$\angle{dbc = 15^\circ}\rightarrow \angle{doc} = 30^\circ\rightarrow EC = OE.$$ $BE = 2C E\rightarrow AD = 2dc$.

**Q.E.D.**

**5, set $ABCD $ is rounded inside the quadrilateral, diagonal $AC $ with $BD $ in $X $, from $X $ for $AB, BC, CD, da$ perpendicular, perpendicular respectively for $A ^\prime, B^\prime, C^\prime, d^\prime$, Verification: $A ^\prime b^\prime + c^\prime d^\prime = a^\prime d^\prime + b^\prime c^\prime$.**

**Prove:**

A known vertical relationship is available $A ^\prime, B, B^\prime, x$ four points, with a diameter of $BX $. Then by the sine theorem there is $${bx\over\sin90^\circ} = {a^\prime b^\prime \over \sin\angle{a^\prime bb^\prime}}\rightarrow {A^\prime B^\ Prime\over BX} = \sin\angle{a^\prime bb^\prime}.$$ on the other hand, in $ABCD $ circumscribed circle also has $${ac\over d} = \sin\angle{a^\prime bb^\prime},$ $ $d of which is $ABCD $ circumscribed circle diameter. The $ $A ^\prime b^\prime = {Bx\cdot Ac\over d}.$$ similarly can be $ $C ^\prime d^\prime = {Dx\cdot Ac\over d}.$$ therefore, $ $A ^\prime B^\prim E + c^\prime d^\prime = bd\cdot{AC \over d}.$$ similarly available $ $B ^\prime c^\prime + a^\prime d^\prime = ac\cdot{BD \over d}.$$ ie $ A^\prime B^\prime + c^\prime d^\prime = a^\prime d^\prime + b^\prime c^\prime$.

**Q.E.D.**

**6, quadrilateral $ABCD $ within the circle, the center of the other circle $O $ on the side $AB $, and tangent to the remaining three sides, verify: $AD + BC = ab$.**

**Prove:**

Considered by the proof conclusion to be $AD, the bc$ translates to $AB $, and the $BE is intercepted on $AB $ = bc$, which proves that the target is $AE = ad$.

by $A, B, C, d$ four points round and tangent properties of $$\angle{bec} = \angle{bce} = {1\over2} (180^\circ-\angle{ebc}) = {1\OVER2}\ANGLE{ADC} = \angl e{odc},$$ therefore $O, E, C, d$ Four points are all round. This can be $$\angle{aed} = \angle{ocd} = {1\over 2}\angle{bcd} = {1\over2} (180^\circ-\angle{a}), $$ is proof $ae = ad$.

**Q.E.D.**

**7, the two circles with each other in the $D $, the line cut a round in $A $, the other round in the $B, C $ Two, verify: $A $ to a straight line $BD, cd$ distance equal.**

**Prove:**

The distance between the points and the two lines (line segments) is equal, which can be linked to the nature of the angular divide line. Extending the $CD $ round in $E $, consider proving $AD $ is the $\angle{bde}$ of the Horn of the split line.

Over point $D $ make two rounds of Gongsche $AC $ $F $, $$\angle{adb} = \ANGLE{ADF} + \angle{bdf} = \angle{daf} +\ANGLE{DCF} = \angle{ade}.$$ that proves $ Ad$ is the $\angle{bde}$ of the horn.

**Q.E.D.**

**Scan attention to "Austrian mathematics Garden" public number (Id:aoshu_xueyuan)**

Math Olympic Question answer: Goal 2017 Junior Maths League training Team homework answer-1