Prove that ∆ABC is an isosceles triangle if the altitude AD from A on BC bisects BC
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in ΔABC , AD is the transversal so AD divides BC into
2 equal halves
so, BD = DC
∠A=∠A [ common ]
AD=AD [ common ]
So , by [SAS] congruence
ΔABD=ΔACD
So,
by [CPCT]
AB=AC
So,
ΔABC is an isosceles triangle
2 equal halves
so, BD = DC
∠A=∠A [ common ]
AD=AD [ common ]
So , by [SAS] congruence
ΔABD=ΔACD
So,
by [CPCT]
AB=AC
So,
ΔABC is an isosceles triangle
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