China
If the altitudes from the two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.
Answer:
- Let ^@BD^@ and ^@CE^@ be the altitudes from the vertices ^@B^@ and ^@D^@ of ^@\triangle ABC.^@
As the altitude of a triangle is perpendicular to the opposite side, we have @^BD \perp AC \text { and } CE \perp AB @^ Also, we are told that the altitudes are of equal length. @^ \implies BD = CE @^ - We need to prove that ^@AB = AC.^@
- In ^@\triangle ADB^@ and ^@\triangle AEC^@, we have @^ \begin{aligned} &BD = CE &&[\text{Given}] \\ &\angle BAD = \angle CAE &&[\text{Common}]\\ &\angle ADB = \angle AEC = 90^ \circ &&[ BD \perp AC \text { and } CE \perp AB] \\ \therefore \space &\triangle ADB \cong \triangle AEC &&[\text{By AAS criterion}] \end{aligned}@^
- As the corresponding parts of congruent triangles are equal, we have
^@AB = AC .^@
Hence, ^@\triangle ABC^@ is isosceles.
