Q26 of 47 Page 10

AB is a diameter of a circle. P is a point on the semi-circle APB. AH and BK are perpendiculars from A and B to the tangent at P. Prove that AH + BK = AB.

Given: AB is a diameter


To prove: AH + BK = AB


Theorem Used:


Tangent to a circle at a point is perpendicular to the radius through the point of contact.


Explanation:



As BK ⊥ HM and AH ⊥ HM and OP⊥ HM.


∴ BK||AH||OP


Let AH = a, BK = b and OP = r, BM = c


In ΔMKB and ΔMHA


∠MKB = ∠MHA = 90°


∠BMK = ∠AMH (common)


By AA criteria







ΔMKB ~ ΔMHA


As triangles are similar ratio of their corresponding sides are also equal.





⇒ 2rb + bc = ac


⇒ 2rb = ac – bc


….. (i)


In ΔMKB and ΔMPO


∠MKB = ∠MPO = 90°


∠BMK = ∠OMP (common)


By AA criteria







ΔMKB ~ ΔMPO


As triangles are similar ratio of their corresponding sides are also equal.





⇒ rb + bc = rc


⇒ rb = rc – bc


…. (ii)


From (i) and (ii),



⇒ 2r – 2b = a – b


⇒ 2r = a + b


⇒ AB = AH + BK


Hence proved.


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