abc is an isosceles triangle with ab=ac draw it seems that every time i draw a perpendicular from some point E on (bc) to meet the extension of (cs) at D, the triangle dax seem to be isosceles . can u help solve my suspicion
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Answer:
IN UP U SEE THE PIC OF THE PROBLEM
Step-by-step explanation:
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)
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