Question

please help me✨
solve this both questions✨
no wrong answer otherwise reported✨
thanks✨
its urgent so answer me fast.✨​

Answer 1

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4 } °°° Solution °°°

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In ΔBCF andΔCBE,

∠BFC=∠CEB (Each 90º)

Hyp. BC= Hyp. BC (Common Side)

Side FC= Side EB (Given)

∴ By R.H.S. criterion of congruence, we have

ΔBCF≅ΔCBE

∴∠FBC=∠ECB (CPCT)

In ΔABC,

∠ABC=∠ACB

[∵∠FBC=∠ECB]

∴AB=AC (Converse of isosceles triangle theorem)

∴ ΔABC is an isosceles triangle...

5 } °°° Solution °°°

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In ΔABP and ΔACP,

∠APB=∠APC (Both equal to 90 0 )

AB=AC [∵ΔABC is an isosceles triangle.]

AP=AP (Common Side)

By R.H.S. criterion of congruence,

△ABP≅△ACP

⇒∠B=∠C (CPCT)

Answer 2

Answer:

rishabh has given you correct answer

it's harshu

bye