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# ABC is an isosceles triangle with AB = AC. Draw AP perpendicular to BC to show that $$\angle{B}$$ = $$\angle{C}$$

Answer :

In $$\triangle{ABP}$$ and $$\triangle{ACP}$$, we have,
AB = AC ...(Given)
AP is the Common side.
$$\angle{ABP}$$ = $$\angle{ACP}$$ = $$90^\circ$$ ...(Since, AP perpendicular to BC)
Therefore, $$\triangle{ABP}$$ $$\displaystyle \cong$$ $$\triangle{ACP}$$ ...(By RHS congruency test)
Thus, $$\angle{B}$$ = $$\angle{C}$$ ...(By CPCT)
Hence, proved.