ICSE Class 10 Answered
Proov
![question image](https://images.topperlearning.com/topper/new-ate/12126247179245460Screenshot2018090712361062.png)
Asked by dr_pradip27121972 | 07 Sep, 2018, 12:40: PM
PART 1:(diagram: not to scale)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/d273b3104bf768393ae92c570292b6e25b934115bfd470.69037313Picture1.png)
OB and OC are the bisectors of exterior angles
construction:
drop perpendiculars
construction:
drop perpendiculars
OD to AB
OE to BC
OE to BC
OF to AC
now perpendiculars from a point on angle bisector to the arms of angle have equal length
so OD = OE
and OE = OF
so
OD = OF
that means perpendicular from O to both the arms of angle BAC is equal
so O should lie on the perpendicular bisector of BAC
Part 2:
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/57848c4d108d56917d3e4d983f2e5e955b934b7bb1b557.19924305Picture2.png)
Here we need to show that AD passes through the midpoint of FE
FO = OE ......OD is median
triangle AOF and triangle ADB
since FE || BC.....mid point theorem
By AA test triangle AOF ~ triangle ADB
so
FO/DB =AF/AB=1/2.......F is midpoint
FO=1/2 DB............a
similarly, we can prove
OE = 1/2 DC.........b
but
DB=DC........D is midpoint
so by a,b and above eqn
we say
FO = OE and OD is median
same with all other sides thus G is same for both the triangles
PART 3:
this is pure understanding
now circumcenter is the point of concurrence of perpendicular bisectors, which are perpendicular
also, orthocenter is the point of concurrence of altitudes, which also are perpendicular
In the diagram, we know AB||ED.....mid point theorem
so perpendicular bisector of AB will be perpendicular to DE
so
the point of concurrence of perpendicular bisectors of ABC = the point of concurrence of altitudes of FED
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/7ba974a86df2974f73f49fdcd8ae15155b934f9025b5f2.27066449Picture3.png)
Answered by Arun | 08 Sep, 2018, 10:01: AM
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