Q5 of 56 Page 134

In ΔABC if ∠B = 90° and BD perpendicular to hypotenuse AC then prove that ΔADB ~ ΔBDC.


Let ∠ABD = x


⇒ ∠BAD = 90 – x


∠DBC = 90 – x


⇒ DCB = 90 – (90 – x) = x


In ΔADB&ΔBDC,


∠ADB = ∠BDC = 90°


∠ABD = ∠DCB = x


∠BAD = ∠DBC = 90 – x


⇒ ΔADB~ΔBDC by AAA Similarity Rule


Hence, proved.


More from this chapter

All 56 →
4

Find the solutions of the following questions:

PB and QA are perpendicular at points B and A of line segment AB. If P and Q lie on opposite sides of AB and on joining P and Q it intersects AB at O and PO = 5 cm, QO = 7 cm, area of ΔPOB = 150 cm2 then find the area of ΔQOA.

 4

Find the solutions of the following questions:

Find the value of x in terms of a, b and c in the figure.


 6

Prove that area of an equilateral triangle formed an one side of a square is half of the area of the equilateral triangle formed on one diagonal of that square itself.

 1

If figure DE || BC. If AD = 4 cm, DB = 6 cm and AE = 5 cm, then the value of EC will be: