Competition problem — Geometry — Maths Olympiad Prep

Library / /395 of 520

Geometry Difficulty 5.7 AIME, harder Prove it

6. As shown in Figure 21, in $\triangle ABC$ , $AB=AC$ , $D$ is the midpoint of $BC$ , and $E$ is any point in $\triangle ABD$ . Connect $AE$ , $BE$ , and $CE$ . Prove: $\angle AEB > \angle AEC$

Solution

(提示: 如图 21, 作点 $E$ 关于 $A D$ 的对称点 $E^{\prime}$ , 联结 $A E^{\prime} 、 C E^{\prime} 、 E E^{\prime}$ , 并延长 $E E^{\prime}$ 交 $A C$ 于点 $F$ . 根据对称性得 $\triangle A B E \cong \triangle A C E^{\prime}$ . 所以, $\angle A E B=\angle A E^{\prime} C$ . 易知 $\angle A E^{\prime} C=\angle A E^{\prime} F+\angle C E^{\prime} F>\angle A E F+\angle C E F=$ $\angle A E C$ , 即 $\angle A E B>\angle A E C$ . )

(提示: As shown in Figure 21, construct the symmetric point $E^{\prime}$ of point $E$ with respect to $A D$ , and connect $A E^{\prime}$ , $C E^{\prime}$ , and $E E^{\prime}$ . Extend $E E^{\prime}$ to intersect $A C$ at point $F$ . According to the symmetry, $\triangle A B E \cong \triangle A C E^{\prime}$ . Therefore, $\angle A E B=\angle A E^{\prime} C$ . It is easy to see that $\angle A E^{\prime} C=\angle A E^{\prime} F+\angle C E^{\prime} F>\angle A E F+\angle C E F=$ $\angle A E C$ , which means $\angle A E B>\angle A E C$ . )

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.