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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Since, < CED = < CAB ( each being 90°)
=> ED // AB
=> < CDE = <CBA ( corresponing angles formed by // lines)
=> tri CED ~ tri CAB ( by AA similarity corollary)
=> CE/CA = ED/AB ( csst)
=> x/ 6 = DE/ 4
But DE = EA ( as tri EAD is isosceles, being each base angle = 45° ) & EA= 6 - x
=> x/6 = ( 6- x) / 4
=> 4x = 36 - 6x
=> 10x = 36
=> x = 3.6
=> DE = 6-x = 6 - 3.6 = 2.4 cm
=> 12/DE = 12/ 2.4 = 5
12/DE = 5 cm . . . . . . .Ans
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