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#