Question

14
Show that PM?=QM - MR.
8.
Show that (1) AB2=BC - BD.
(ii) AC2=BC - DC
(iii) AD2= BD.CD.
6.
5. ABD is a triangle right angled at A and AC I BD
e right angled at C. Prove that AB2=2AC2.​

Answer 1

Step-by-step explanation:

(i) In ΔADB and ΔCAB, we have

∠DAB = ∠ACB (Each equals to 90°)

∠ABD = ∠CBA (Common angle)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB2 = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,

∠CBA = 180° - 90° - x

∠CBA = 90° - x

Similarly, in ΔCAD

∠CAD = 90° - ∠CBA

           = 90° - x

∠CDA = 180° - 90° - (90° - x)

∠CDA = x

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each equals to 90°)

∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC2 =  DC × BC

(iii) In ΔDCA and ΔDAB, we have

∠DCA = ∠DAB (Each equals to 90°)

∠CDA = ∠ADB (common angle)

∴ ΔDCA ~ ΔDAB [By AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD2 = BD × CD

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