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Triangle ABC and triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the
same side of BC (see Figure). If AD is extended to intersect BC at P, show that
(i) triangleABD ≅ triangle ACD
(ii) triangle ABP ≅ triangle ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
(i) In ΔABD and ΔACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common)
∴ ΔABD ≅ ΔACD (By SSS congruence rule)
⇒ ∠BAD = ∠CAD (By CPCT)
⇒ ∠BAP = ∠CAP …. (1)
(ii) In ΔABP and ΔACP,
AB = AC (Given)
∠BAP = ∠CAP [From equation (1)]
AP = AP (Common)
∴ ΔABP ≅ ΔACP (By SAS congruence rule)
⇒ BP = CP (By CPCT) … (2)
(iii) From equation (1),
∠BAP = ∠CAP
Hence, AP bisects ∠A.
In ΔBDP and ΔCDP,
BD = CD (Given)
DP = DP (Common)
BP = CP [From equation (2)]
∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule)
⇒ ∠BDP = ∠CDP (By CPCT) … (3)
Hence, AP bisects ∠D.
(iv) ΔBDP ≅ ΔCDP
∴ ∠BPD = ∠CPD (By CPCT) …. (4)
∠BPD + ∠CPD = 180
(Linear pair angles)
∠BPD + ∠BPD = 180
2∠BPD = 180 [From equation (4)]
∠BPD = 90 … (5)
From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠ A.
(i) In ΔBAD and ΔCAD,
∠ADB = ∠ADC (Each 90º as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
∴ΔBAD ≅ ΔCAD (By RHS Congruence rule)
⇒ BD = CD (By CPCT)
Hence, AD bisects BC.
(ii) Also, by CPCT,
∠BAD = ∠CAD
Hence, AD bisects ∠A.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of â?? PQR (see Figure). Show that:
(i) triangle ABM ≅ triangle PQN
(ii) triangle ABC ≅ triangle PQR
(i) In ΔABC, AM is the median to BC.
∴ BM = 
BC
In ΔPQR, PN is the median to QR.
∴ QN = QR
However, BC = QR
∴ BC = QR
⇒ BM = QN … (1)
In ΔABM and ΔPQN,
AB = PQ (Given)
BM = QN [From equation (1)]
AM = PN (Given)
∴ ΔABM ≅ ΔPQN (SSS congruence rule)
∠ABM = ∠PQN (By CPCT)
∠ABC = ∠PQR … (2)
(ii) In ΔABC and ΔPQR,
AB = PQ (Given)
∠ABC = ∠PQR [From equation (2)]
BC = QR (Given)
⇒ ΔABC ≅ ΔPQR (By SAS congruence rule)
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
In ΔBEC and ΔCFB,
∠BEC = ∠CFB (Each 90°)
BC = CB (Common)
BE = CF (Given)
∴ ΔBEC ≅ ΔCFB (By RHS congruency)
⇒ ∠BCE = ∠CBF (By CPCT)
∴ AB = AC (Sides opposite to equal angles of a triangle are equal)
Hence, ΔABC is isosceles.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
In ΔAPB and ΔAPC,
∠APB = ∠APC (Each 90º)
AB =AC (Given)
AP = AP (Common)
∴ ΔAPB ≅ ΔAPC (Using RHS congruence rule)
⇒ ∠B = ∠C (By using CPCT)
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