In a triangle ABC, AB = AC and D is a point on side AC such that BC 2 = AC × CD. Prove that BD = BC.

Draw BE ⊥ AC. In right triangle BEC, By pythagoras theorem, ⇒ BC 2 = BE 2 + EC 2 ⇒ BC 2 = BE 2 + (AC - AE) 2 ⇒ BC 2 = BE 2 + AC 2 + AE 2 - 2AC.AE .........(i) In right triangle ABE, By pythagoras theorem, ⇒ AB 2 = BE 2 + AE 2 .........(ii) Substituting value of BE 2 + AE 2 from (ii) in (i) we get, ⇒ BC 2 = AB 2 + AC 2 - 2AC.CE ⇒ BC 2 = AC 2 + AC 2 - 2AC.AE [∵ AB = AC] ⇒ BC 2 = 2AC 2 - 2AC.AE ⇒ BC 2 = 2AC(AC - AE) ⇒ BC 2 = 2AC × EC .........(iii) Given, BC 2 = AC × CD ..........(iv) Comparing (iii) and (iv) we get, ⇒ 2AC × EC = AC × CD ⇒ 2EC = CD or EC = So, we can say that ED = EC as E is mid-point of CD. In △BDE and △BCE, BE = BE (Common) ED = EC ∠BED = ∠BEC = 90° △BDE ≅ △BCE by SAS axiom of congruency. We know that corresponding parts of congruent triangle are equal. ∴ BD = BC. Hence, proved that BD = BC.

In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², Prove that ∠ACD = 90°.

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In the adjoining figure, find the length of AD in terms of b and c.

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ABCD is a square, F is mid-point of AB and BE is one third of BC. If area of △FBE is 108 cm², find the length of AC.

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In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that
8PT² = 3PR² + 5PS².

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Mathematics

In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC × CD.
Prove that BD = BC.

Pythagoras Theorem

Related Questions

In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², Prove that ∠ACD = 90°.

In the adjoining figure, find the length of AD in terms of b and c.

ABCD is a square, F is mid-point of AB and BE is one third of BC. If area of △FBE is 108 cm², find the length of AC.

In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that
8PT² = 3PR² + 5PS².

Mathematics

In a triangle ABC, AB = AC and D is a point on side AC such that BC2 = AC × CD.Prove that BD = BC.

Pythagoras Theorem

Related Questions

In a quadrilateral ABCD, ∠B = 90°. If AD2 = AB2 + BC2 + CD2, Prove that ∠ACD = 90°.

In the adjoining figure, find the length of AD in terms of b and c.

ABCD is a square, F is mid-point of AB and BE is one third of BC. If area of △FBE is 108 cm2, find the length of AC.

In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that8PT2 = 3PR2 + 5PS2.

In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC × CD.
Prove that BD = BC.

In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², Prove that ∠ACD = 90°.

ABCD is a square, F is mid-point of AB and BE is one third of BC. If area of △FBE is 108 cm², find the length of AC.

In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that
8PT² = 3PR² + 5PS².