Advertisements
Advertisements
प्रश्न
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
Advertisements
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
APPEARS IN
संबंधित प्रश्न
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that
- ΔABD ≅ ΔBAC
- BD = AC
- ∠ABD = ∠BAC.
Which congruence criterion do you use in the following?
Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
Which of the following statements are true (T) and which are false (F):
If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
Which of the following statements are true (T) and which are false (F):
Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB = CE.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that: QA = QB.
In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥ PQ;
Prove that:
- ΔXTQ ≅ ΔXSQ.
- PX bisects angle P.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
