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Answer:
Given: ∆ABC with the altitudes from vertex B and C intersect at point M, so that BM = CM.
To prove:∆ABC is isosceles
Proof:-Let the altitudes from vertex B intersects AB at D and from C intersects AC at E( with reference to the figure)
Consider ΔBMC where BM=MC
Then ∠CBM=∠MCB......(1)(Angles opposite to equal sides of a triangle are equal)
Now Consider ΔDMB and ΔCME
∠D=∠E.......(each 90°)
BM=MC...............(given)
∠CME=∠BMD........(vertically opposite angles)
So by ASA congruency criteria
ΔDMB ≅ ΔCME
∴∠DBM=∠MCE........(2)(corresponding parts of a congruent triangle are equal)
Adding (1) and (2),we get
∠DBM+∠CBM=∠MCB+∠MCE
⇒∠DBC=∠BCE
⇒∠B=∠C⇒AB=AC(sides opposite to equal angles of a triangle are equal)⇒∆ABC is an isosceles triangle .
