Question

In an isosceles triangle, ABC, AB =AC and BC is produced to D. prove that
$ad {}^{2} - ac {}^{2} = bc \times cd$
​

Answer 1

Given: In isosceles triangle, AB = AC and BC is produced to D.

To Prove: AD² - AC² = BD × CD

To Construct: AE as an altitude and since here it's an isosceles ∆ therefore a median too.

Proof: By applying Pythagoras Theorem.

In ∆AED, we have

→ AD² = AE² + DE²

→ AE² = AD² - DE²

In ∆AEC, we have

→ AC² = AE² + CE²

→ AE² = AC² - CE²

By equating we get,

→ AD² - DE² = AC² - CE²

→ AD² - AC² = DE² - CE²

→ AD² - AC² = (DE + CE)(DE - CE)

→ AD² - AC² = (DE + BE)(DE - CE)

→ AD² - AC² = BD · CD

Q.E.D

Answer 2

Correct Question—

In an isosceles triangle, ABC, AB =AC and BC is produced to D. prove that

$AD {}^{2} - AC {}^{2} = BD \times CD$

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Given—

ABC is an isosceles Triangle

AB = AC

BC is produced to D

To prove—

AD² - AC ² = BD × CD

Let Construct something —

AE is constructed as altitude and we also know that altitude of isosceles triangle is median also.

Proof:-

In ΔAED

using Pythagorous

AD² = AE² + DE²

⇒AE² = AD² - DE² ______(1)

In ΔAEC

again by Pythagorous

AC² = AE² + CE²

⇒AE² = AC² - CE² _______(2)

NOW,

We can equate (1) & (2)

AD² - DE² = AC² - CE²

⇒AD² - AC² = DE² - CE²

Breaking by algebric formula

⇒AD² - AC² = (DE - CE)(DE + CE)

As CE = BE by Median LAW

⇒AD² - AC² = (DE - CE)(DE + BE)

⇒AD² - AC² = CD × BD

⇒AD² - AC² = BD × CD

Hence, Proved

$\huge{\red{\ddot{\smile}}}$