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In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Figure). Show that â?? ABC ≅ â?? ABD. What can you say about BC and BD?
In ΔABC and ΔABD,
AC = AD (Given)
∠CAB = ∠DAB (AB bisects ∠A)
AB = AB (Common)
∴ ΔABC ≅ ΔABD (By SAS congruence rule)
∴ BC = BD (By CPCT)
Therefore, BC and BD are of equal lengths.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fiure). Prove that
(i) â?? ABD ≅ â?? BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
In ΔABD and ΔBAC,
AD = BC
∠DAB = ∠CBA
AB = AB (common side)
∴ ΔABD ≅ ΔBAC by SAS Test of congruency
∴ BD = AC (corressponding sides of congruent traingles)
∴ ∠ABD = ∠BAC (corressponding angles of congruent traingles)
AD and BC are equal perpendiculars to a line segment AB (see Figure). Show that CD bisects AB.
In ΔOCB and ΔODA,
BC = AD
∠CBO = ∠ DAO (both are 90°)
∠COB = ∠ DOA (vertically opposite angles)
∴ΔOCB ≅ ΔODA by AAS Test of congruency
∴ OB = OA (corressponding sides of congruent triangles)
∴ CD bisects AB
l and m are two parallel lines intersected by another pair of parallel lines p and q (see Figure). Show that â?? ABC ≅ â?? CDA.
In ΔABC and ΔCDA
∠BAC = ∠DCA (alternate interior angles)
∠BCA = ∠DAC (alternate interior angles)
AC = AC (common side)
∴ ΔABC ≅ ΔCDA by ASA Test of congruency
In Figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
∠ BAD = ∠ EAC
∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∴ ∠ BAC = ∠ EAD
Now, in ΔABC and ΔADE
AB = AD
AC = AE
∠ BAC = ∠ EAD
∴ ΔABC ≅ ΔADE by ASA Test of congruency
∴ BC = DE (corressponding sides of congruent triangles)
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Figure). Show that
(i) â?? DAP ≅ â?? EBP
(ii) AD = BE
AB is a line segment and P is its mid-point. So, AP = PB
∠ EPA = ∠ DPB
∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∴ ∠ APD = ∠ BPE
In ΔDAP and ΔEBP,
∠ PAD = ∠ PBE (Given, ∠ BAD = ∠ ABE)
AP = PB
∠ APD = ∠ BPE
∴ ΔDAP ≅ ΔEBP by ASA Test of congruency
∴ AD = BE (Corressponding sides of congruent triangles)
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B
(see Figure). Show that:
(i) â?? AMC ≅ â?? BMD
(ii) ∠ DBC is a right angle.
(iii) â?? DBC ≅ â?? ACB
(iv)
(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point of hypotenuse AB)
∠AMC = ∠BMD (vertically opposite angles)
CM = DM
ΔAMC ≅ ΔBMD by SAS Test of congruency
(ii) ∠ACM = ∠BDM (corressponding angles of congruent triangles)
Also, ∠ACM and ∠BDM form alternate interior angles with CD as the transversal.
So, AC parallel to BD.
So, ∠ACB and ∠DBC form co-interior angle with BC as transversal.
So, ∠ACB + ∠DBC = 180° (co-interior angle adds up to 180°)
∠ACB = 90° (It is given that in triangle ABC, right angled at C)
So, ∠DBC = 90°. So, ∠DBC is a right angle.
(iii) ΔAMC ≅ ΔBMD by SAS Test of congruency
∴ AC = BD (corressponding sides of congruent triangles)
Now, in ΔDBC and ΔACB
BD = AC
∠DBC = ∠ACB (both are 90°)
BC = BC (common side)
∴ ΔDBC ≅ ΔACB by SAS Test of congruency
(iv) ∴ DC = AB (corressponding sides of congruent triangles)
∴ DM + MC = AB
∴ MC + MC = AB (It is given, DM = MC)
∴ 2 MC = AB
∴ MC = 1/2 AB
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 Figure). Show that:
(i) Triangle APB ≅ triangle AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.
(i) ∠APB = ∠AQB (It is given that both the angles are 90°)
(ii) In ΔABP and ΔABQ, ∠APB = ∠AQB (both are 90°)
∠BAP = ∠BAQ (line l bisects ∠A into ∠BAP and ∠BAQ)
Side BA ≅ Side BA (common side of both the triangles)
Hence, ΔABP ≅ ΔABQ by AAS Test.
Hence, BP = BQ (corressponding sides of congruent triangles)
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