Question #0583a

2 Answers
Mar 14, 2017

see explanation.

Explanation:

enter image source here

Given that angleABC=90^@,
=> AC is the diameter of the circumscribed circle of DeltaABC.
As BE is perpendicular to diameter AC,
=> BD=DE

Intersecting chord theorem : when two chords intersect each other inside a circle, the products of their segment are equal.
=> AD*DC=BD*DE=BD^2
=> 16*9=BD^2, => BD=12.

By Pythagorean Theorem:
x^2=AD^2+BD^2=16^2+12^2=400
=> x=sqrt400=20

y^2=CD^2+BD^2=9^2+12^2=225
=> y=sqrt225=15

Hence, BD=12, x=20, y=15

Mar 15, 2017

Sol. 2. (see explanation).

Explanation:

enter image source here

Solution 2.

As kindly suggested by Ratnaker Mehta, the following results are very useful in such geometrical problems.

(1) BD^2=AD*DC
(2) AB^2=AD*AC
(3) CB^2=CD*CA

=> BD^2=16*9=144, => BD=sqrt144=12
=> AB^2=16*(16+9)=400, => AB=sqrt400=20
=> CB^2=9*(16+9)=225, => CB=sqrt225=15

proof :

DeltaABC, DeltaADB and DeltaBDC are similar triangles.

(1) (AD)/(DB)=(BD)/(DC), => BD^2=AD*DC

(2) (AB)/(AC)=(AD)/(AB), => AB^2=AD*AC

(3) (BC)/(CA)=(DC)/(CB), => BC^2=CD*CA