In Fig 6.13, OB is the perpendicular bisector of the line segment DE, FA ⊥ OB and FE intersects OB at the point C. Prove that 1/OA + 1/OB = 2/OC
Solution:
Given, OB is the perpendicular bisector of the line segment DE.
FE intersects OB at point C.
Also, FA ⊥ OB
We have to prove that 1/OA + 1/OB = 2/OC.
In △OFA and △ODB,
∠O = ∠O = common angle
∠A = ∠B = 90°
AAA criterion states that if two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angle will also be equal.
By AAA criterion, the third angle will be equal.
Therefore, the triangles OFA and ODB are similar.
By the property of similar triangles,
The corresponding sides are proportional.
OA/OB = AF/DB -------------- (1)
In △FCA and △ECB,
∠A = ∠B = 90°
Vertically opposite angles are equal.
i.e., ∠FCA = ∠ECB
By AAA criterion, the third angle will be equal.
Therefore, the triangles FCA and ECB are similar.
By the property of similar triangles,
The corresponding sides are proportional.
FA/BE = AC/BC -------------- (2)
From the figure,
B is the midpoint of DE.
So, BE = BD
Substitute the BE in (2),
FA/BD = AC/BC ---------------- (3)
Equating (1) and (2),
OA/OB = AC/BC ---------------- (4)
From the figure,
AC = OC - OA
BC = OB - OC
Substituting the values of AC and BC in (4),
OA/OB = (OC - OA)/(OB - OC)
OA(OB - OC) = OB(OC - OA)
OA.OB - OA.OC = OB.OC - OA.OB
Or rearranging,
OA.OB + OA.OB = OB.OC + OA.OC
2(OA.OB) = OB.OC + OA.OC
Dividing by OA.OB.OC on both sides,
2/OC = 1/OA + 1/OB
Therefore, it is proved that 1/OA + 1/OB = 2/OC
✦ Try This: In ΔABC, if AD is the bisector of ∠A, prove that Area (ΔABD)/Area (ΔACD) =AB/AC
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.4 Sample Problem 1
In Fig 6.13, OB is the perpendicular bisector of the line segment DE, FA ⊥ OB and FE intersects OB at the point C. Prove that 1/OA + 1/OB = 2/OC
Summary:
In Fig 6.13, OB is the perpendicular bisector of the line segment DE, FA ⊥ OB, and FE intersects OB at point C. It is proven that 1/OA + 1/OB = 2/OC
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