5
ACROSS
4. objects that have no connection to the source file
5. view which shows a window that splits into slide
pane, notes pane and a left pane having slides/
outline tab
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Answer:
Lets draw altitude from A to BC such that it meets BC at D, angles are ΔADC,∠D=90
o
⇒AD
2
+DC
2
=AC
2
⇒AD
2
+(BD+BC)
2
=AC
2
⇒AD
2
+BD
2
+BC
2
+2BD.BC=AC
2
………….(1)
In ΔADB,∠D=90
o
⇒AD
2
+BD
2
=AB
2
Substituting in equation (1)
⇒AD
2
+BD
2
+BC
2
+2BD.BC=AC
2
⇒AB
2
+BC
2
+2BD.BC=AC
2
as in ΔADB,∠DAB=∠45
o
⇒AD=DB(opposite sides are equal)
⇒AB
2
+BC
2
+2BD×BC=AC
2
⇒AB
2
+BC
2
+2AD×BC=AC
2
⇒AB
2
+BC
2
+4[
2
1
AD×BC]=AC
2
⇒AB
2
+BC
2
+4(ΔABC)=AC
2
∴ Hence proved.
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