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प्रश्न
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
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उत्तर
Given that, if perpendicular from any point within, an angle on its arms is congruent, prove that it lies on the bisector of that angle
Now,
Let us consider an angle ABC and let BP be one of the arm within the angle
Draw perpendicular PN and PM on the arms BC and BA such that they meet BC and BA in N and M respectively.
Now, in ΔBPM and ΔBPN
We have ∠BMP= BNP = 90° [given]
BP=BP [Common side]
And MP=NP [given]
So, by RHS congruence criterion, we have
ΔBPM≅ΔBPN
Now,
∠MBP=∠NBP [ Corresponding parts of congruent triangles are equal]
⇒ BP is the angular bisector of ∠ABC
∴ Hence proved
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संबंधित प्रश्न
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Which congruence criterion do you use in the following?
Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
Which congruence criterion do you use in the following?
Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
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In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that : ED = EF
In the following diagram, AP and BQ are equal and parallel to each other. 
Prove that: AB and PQ bisect each other.
