In a triangle ABC, AB=AC and D is a point on side AC such that BCxBC=ACxCD. How to prove that BD=BC?

Question

Maverick Smith · Accepted Answer

Given

#"In "DeltaABC#
#AB=AC and D" is a point on"AC " such that"#
#BCxxBC=ACxxAD#
#"We are to prove "BD=BC#
Proof

Rearrenging the given relation

#BCxxBC=ACxxAD" "# We can write

#(BC)/(CD)=(AC)/(BC)->DeltaABC" similar "DeltaBDC#
Their corresponding angle pairs are:
#1. /_BAC "= corresponding "/_DBC#
#2. /_ABC "= corresponding "/_BDC#
#3. /_ACB " =corresponding "/_DCB#
So as per above relation 2 we have
#/_ABC =" corresponding "/_BDC#
#"Again in"DeltaABC#
#AB=AC->/_ABC=/_ACB=/_DCB#
#:."In "DeltaBDC,/_BDC=/_BCD#
#->BD=BC#
Alternative way

The ratio of corresponding sides may be written in extended way as follows
#(BC)/(CD)=(AC)/(BC)=(AB)/(BD)#
From this relation we have
#(AC)/(BC)=(AB)/(BD)#
#=>(AC)/(BC)=(AC)/(BD)->"As "AB=AC" given"#
#=>1/(BC)=1/(BD)#
#=>BC=BD#
Proved

Hope, this will help

Sophia Alfred · Answer

undefined Given triangle $ABC$ with $AB = AC$ and a point $D$ on side $AC$ such that $BC \cdot BC = AC \cdot CD$, we want to prove that $BD = BC$.

Since $AB = AC$, triangle $ABC$ is isosceles. Let $E$ be the midpoint of $BC$.

Now, from the given condition, $BC \cdot BC = AC \cdot CD$, implying that $BC^2 = AC \cdot CD$. Since $AB = AC$, we can rewrite this as $BC^2 = AB \cdot CD$.

Using the Power of a Point Theorem, we can write $BD \cdot CD = AD \cdot DC$.

Since triangle $ABC$ is isosceles, $AE$ is perpendicular to $BC$, so $AE$ is also the median, altitude, and angle bisector of $ABC$. Thus, triangles $ABE$ and $ACE$ are congruent by the hypotenuse-leg congruence criterion. Hence, $BE = EC$.

Now, consider triangle $BDE$. It is a right triangle with $BE = EC$, so $BD = DE$.

Considering triangle $CDE$, we have $CD = CD$, and since $BD = DE$, by the Hypotenuse-Leg congruence criterion, triangles $BDE$ and $CDE$ are congruent. This implies $BC = CE$.

Now, since $BC = CE$ and $BE = EC$, we have $BC = BE + EC = BD$.

Hence, $BD = BC$.