In ∆abc, the altitudes from vertex b and c intersect at point m, so that bm = cm. prove that ∆abc is isosceles.

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 .

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

if x = 1, then the answer is 3.5