Math olympiad question: Plane Geometry-9 (goal 2017 national Junior maths competition class 14th week homework)

1. Set the $D, e$ is $\triangle{abc}$ side $BC $ two, and $BD = ec$, $\angle{bad} = \angle{eac}$. Verification: $\triangle{abc}$ is isosceles triangle. Russia

HINT:

by $BD = ec$ and collinear, considering panning $\triangle{abd}$ to $\triangle{fec}$, you get $ABEF $ is parallelogram and $A, E, C, f$ four points round $\rightarrow \angle{ace} = \angle{afe} = \angle{abd}$.

2. Set $P $ is parallelogram $ABCD $ within any point, and $\ANGLE{APB} + \ANGLE{CPD} = 180^{\circ}$. Proof: $\ANGLE{PBC} = \angle{pdc}$. Canada

HINT:

Consider panning the $\triangle{pab}$ to $\triangle{p^\prime DC} \rightarrow app^\prime D, pbcp^\prime$ are parallelogram $\rightarrow \angle{PCP^\ Prime} + p^\prime DP = 180^\circ \rightarrow P, C, P^\prime, d$ four points total round

$\rightarrow \ANGLE{PDC} = \angle{pp^\prime C} = \angle{pbc}$.

3. Set the quadrilateral $ABCD $ within a little $P $, making the quadrilateral $ABPD $ parallelogram. Proof: if $\ANGLE{CBP} = \angle{cdp}$, then $\ANGLE{ACD} = \angle{bcp}$. Russia

HINT:

Consider panning $\triangle{acd}$ to $\TRIANGLE{BQP} \rightarrow DPQC, abqc$ are parallelogram $\rightarrow \angle{cbp} = \angle{cdp} = \angle{P QC} \rightarrow P, B, Q, C $ four total round

$\rightarrow \angle{bcp} = \angle{bqp} = \angle{acd}$.

4. Set convex hexagonal $ABCDEF $ for all internal angles equal. Proof: $| ab-de| = | Bc-ef| = | cd-fa|$. Russia

HINT:

Respectively over $B, D, f$ as $AF, EF, de$ parallel to the $P, Q, R $, easy license $EF \parallel BC, Cd\parallel AF, Ab\parallel de$ and $\triangle{pqr}$ is the triangle Shape $\rightarrow PQ = | ab-de|, PR = | cd-af|, RQ = | bc-ef|$.

5. In $\triangle{abc}$, $AB = ac$, $\angle{a} = 20^{\circ}$, $D, e$ are edge $AC, points on ab$, make $\angle{dbc} = 60^{\circ}$, $\an GLE{ECB} = 50^{\circ}$. The degree of seeking $\angle{bde}$.

HINT:

Note that $\triangle{abd}, \triangle{ebc}$ are isosceles triangle, desire $\angle{bde}$ degrees can be considered to find $\angle{aed}$ degrees. by $\angle{ace} = 30^\circ$, made $E $ about $AC $ symmetric point $E ^\prime$, easy to license $\triangle{aee^\prime}$ is isosceles triangle and $\triangle{ece^\prime}$ is a positive triangle . $\because \angle{abd} = \angle{dbe^\prime} = 20^\circ$ and $AD $ split $\angle{eae^\prime} \rightarrow d$ is $\triangle{ABE^\prim e}$ inside Heart $\rightarrow \angle{ae^\prime D} = 50^\circ \rightarrow \angle{aed} = 50^\circ \rightarrow \angle{BDE} = 30^\circ$ .

6. The three sides of the $\triangle{abc}$ are $a, B, C $, respectively, the vertices of which are $A, B, C $, and $AB $ on the edge of $h $, proving: $a +b \ge \sqrt{c^2 + 4h^2}$.

HINT:

Consider constructing a $\sqrt{c^2 + 4h^2}$. Over $C $ for $l \parallel ab$, making $A $ $l the symmetry point $A ^\prime$. Easy License $A ^\prime B = \sqrt{c^2 + 4h^2} \le a^\prime C + BC = A + b$.

7. Isosceles $\triangle{abc}$ 's two-waist $AB, ac$ on the points $D, e$, so that $AD = ae$. Respectively over $A, d$ for $BE $ of the vertical margin $BC $ $M, n$. Proof: $M $ for the midpoint of $NC $ when and only if $AC \perp ab$.

HINT:

If $AC \perp ab$, make $D $ on the symmetry point of $AC $ $D ^\prime$, link $DC, D^\prime C $. Easy-to-know $\angle{d^\prime CA} = \ANGLE{DCA} = \angle{abe} = \angle{eam} \rightarrow d^\prime C \parallel AM \parallel DN$, while $A $ is $DD ^\prime$ midpoint $\rightarrow NM = mc$.

If $M $ is $NC $ among points, times the length $DA $ to $D ^\prime \rightarrow DN \parallel AM \parallel d^\prime C $, Nexus $DE, D^\prime e$. Easy-to-know $EA = AD = Ad^\prime \rightarrow DE \perp d^\prime e$ and $BE \perp d^\prime C \rightarrow e$ is $\triangle{d^\prime BC}$ The vertical heart $\rightarrow AC \perp ab$.

8. In convex quadrilateral $ABCD $, $\angle{abc} = 30^{\circ}$, $\angle{adc} = 60^{\circ}$, $AD = ac$. Proof: $BD ^2 = ab^2 + bc^2$.

HINT:

Consider constructing right triangle. Rotate the $\triangle{cdb}$ around $C $ point counterclockwise $60^\circ$ to $\triangle{cab^\prime}$. Yi $\triangle{cbb^\prime}$ is a positive triangle $\rightarrow bd^2 = {ab^\prime}^2 = ab^2 + {bb^\prime}^2 = ab^2 + bc^2$.

9. Set $P $ is square $ABCD $ inside a bit, so $\angle{pad} = \angle{pda} = 15^{\circ}$. Proof: $\triangle{pbc}$ is equilateral triangle.

HINT:

Note that if the conclusion is established, then $\TRIANGLE{ABP}, \triangle{cdp}$ is the apex of the $30^\circ$ isosceles triangle, so consider moving $\triangle{apd}$ to $\triangle{abp}$.

The $\triangle{apd}$ is folded to $\triangle{aqb}$ $AC $ for the axis of symmetry, linked $QP $. Yi $\triangle{aqp}$ is a positive triangle $\rightarrow \TRIANGLE{APD} \cong \triangle{pqb} \rightarrow PB = AD = bc$, the same as the certificate $PC = bc$.

$\triangle{abc}$, $\angle{bac} = 40^{\circ}$, $\angle{abc} = 60^{\circ}$, $D $ and $E $ respectively are sides $AC $ and $AB $ on the point, making $\angle {CBD} = 40^{\circ}$, $\ANGLE{BCE} = 70^{\circ}$, $F $ is the intersection of $BD $ and $CE $. Verification: $AF \perp bc$. Canada

HINT:

Refer to the solution of another blog post.

Math olympiad question: Plane Geometry-9 (goal 2017 national Junior maths competition class 14th week homework)