In adjacent figure- sides AB and AC of -ABC are extended to points P and Q respectively- Also- -PBC AB-

Since ∠ PCB lt; ∠ QCB ∠ PBC gt; ∠ QCB Add 180 ^∘ #160; , we get #160; 180 ^∘ #160; ∠ PBC gt; 180 ^∘ #160; ∠ QCB ∠ ABC ...

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Standard IX

Mathematics

AAS Criteria for Congruency

In adjacent f...

Question

In adjacent figure, sides AB and AC of $Δ$ ABC are extended to points P and Q respectively. Also, $∠$ PBC $< ∠$ QCB. Show that AC $>$ AB.

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Solution

Since $∠ P C B < ∠ Q C B$
$- ∠ P B C > - ∠ Q C B$
Add $180^{\circ}$ , we get
$180^{\circ} - ∠ P B C > 180^{\circ} - ∠ Q C B ∠ A B C > ∠ A C B$
Using triangle property , we get
$A C > A B$

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Similar questions

Q. In Figure, sides AB and AC of

Δ

ABC are extended to points P and Q respectively.

Also

∠

PBC

< ∠

QCB. Show that AC

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respectively.
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∠ P B C < ∠ Q C B

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Q. In the given figure sides AB and AC of

Δ

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∠ P B C < ∠ Q C B

. Show that AC > AB.

Question 2
In the given figure sides AB and AC of $Δ$ ABC are extended to points P and Q respectively. Also, $∠ P B C < ∠ Q C B$ . Show that AC > AB.

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In adjacent figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC <∠QCB. Show that AC > AB.

Since ∠PCB<∠QCB−∠PBC>−∠QCBAdd 180∘ , we get 180∘−∠PBC>180∘−∠QCB∠ABC>∠ACBUsing triangle property , we get AC>AB

In adjacent figure, sides AB and AC of $Δ$ ABC are extended to points P and Q respectively. Also, $∠$ PBC $< ∠$ QCB. Show that AC $>$ AB.

Since $∠ P C B < ∠ Q C B$
$- ∠ P B C > - ∠ Q C B$
Add $180^{\circ}$ , we get
$180^{\circ} - ∠ P B C > 180^{\circ} - ∠ Q C B ∠ A B C > ∠ A C B$
Using triangle property , we get
$A C > A B$