0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

SANGUES 8. In Fig 6.34.0 6.54.O is a point in the interior of a triangle COD IBC, OE LAC and OF LAB.Show that A2+OB:+OC-OD--OF-OF=AF+BD+CE. f) AF+BD + CE-=AE+CD+BF: 6 Aladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall. In attached to a vertical pole of height 18 m tached to the other Fig. 6.54

Solution
Verified by Toppr


Was this answer helpful?
0
Similar Questions
Q1
In Fig., O is a point in the interior of a triangle ABC,ODBC,OEAC and OFAB. Show that
(i)OA2+OB2+OC2OD2OE2OF2=AF2+BD2+CE2,
(ii) AF2+BD2+CE2=AE2+CD2+BF2.
785673_bbc6c3a95868420bba0f8996a1d7ff45.png
View Solution
Q2
In the above figure, O is a point in the interior of a triangle ABC, ODBC,OEAC and OFAB. Show that:
(i) OA2+OB2+OC2OD2OE2OF2=AF2+BD2+CE2
(ii) AF2+BD2+CE2=AE2+CD2+BF2
465464.PNG
View Solution
Q3
From a point O in the interior of triangle ABC, perpendiculars OD, OE and OF are drawn to the sides BC,CA and AB respectively. Prove that- (i) AF^2 + BD^2 + CE^2 = OA^2 +OB^2 + OC^2 - OD^2 - OE^2 - OF^2. (ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2
View Solution
Q4
In given figure from a point O in the interior of a ABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :

(i) AF2+BD2+CE2=OA2+OB2+OC2OD2OE2OF2

(ii)AF2+BD2+CE2=AE2+CD2+BF2

1009933_03a895b6b2e54197b5f792881095eb8a.png
View Solution
Q5
From a point O in the interior of a ΔABC. Perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which one of the following is true?
View Solution