D is any point on the side AC of- ABC-with AB - AC- Show that-CD-BD-

Given: In ∆ ABC, AB = AC To prove: CD lt; BD Proof: In ∆ ABC, Since, AB = AC So, ∠ABC=∠ACB (Given) ...(i) In ∆ABC and ∆DBC,∠ ABC gt;∠DBC⇒ ∠ACB gt;∠ DBC F ...

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Byju's Answer

Standard IX

Mathematics

Relationship between Unequal Sides of Triangle and the Angles Opposite to It.

D is any poin...

Question

$D is any point on the side AC of Δ A B C with AB = AC. Show that C D < B D .$

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Solution

Given: In ∆ ABC, AB = AC
To prove: CD < BD
Proof:
In ∆ ABC,
Since, AB = AC
So, ∠ABC=∠ACB
(Given)
...(i)
In ∆ABC and ∆DBC,∠ ABC>∠DBC⇒ ∠ACB>∠ DBC
From i⇒ BD>CD
Side opposite to greater angle is longer.∴ CD<BD

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D is any point on the side AC of ΔABC with AB = AC. Show that CD<BD.

Given: In ∆ ABC, AB = AC To prove: CD < BD Proof: In ∆ ABC, Since, AB = AC So, ∠ABC=∠ACB (Given) ...(i) In ∆ABC and ∆DBC,∠ ABC>∠DBC⇒ ∠ACB>∠ DBC From i⇒ BD>CD Side opposite to greater angle is longer.∴ CD<BD

$D is any point on the side AC of Δ A B C with AB = AC. Show that C D < B D .$

Given: In ∆ ABC, AB = AC
To prove: CD < BD
Proof:
In ∆ ABC,
Since, AB = AC
So, ∠ABC=∠ACB
(Given)
...(i)
In ∆ABC and ∆DBC,∠ ABC>∠DBC⇒ ∠ACB>∠ DBC
From i⇒ BD>CD
Side opposite to greater angle is longer.∴ CD<BD