In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE*(AB+AC)= AB*AC

Question

Ashutosh nayak · Answer

First of all draw the figure.then draw a perpendicular DG to AB.then see by ASA congruence triangleDEA=triangleDGA.so DG=DE.so area of ABC=areaDCA+areaDAB.so .5*AB*AC=.5*AC*DE+.5*AB*DG soo AB*AC=DE*(AB+AC) as DE=DG

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In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE(AB+AC)= ABAC

In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE(AB+AC)= ABAC

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In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE*(AB+AC)= AB*AC

In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE*(AB+AC)= AB*AC

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In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE(AB+AC)= ABAC

In a triangle ABC angle BAC is 90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DE(AB+AC)= ABAC